How does ax^2+bx+c=0 become the quadratic formula?

What method is used to get the quadratic formula from ax^2+bx+c=0 ?

How does ax^2+bx+c=0 become the quadratic formula?
Completing the square.
Reply:ax^2 +bx+c=0





a=1


b=1


c=1





apply it to the quadratic formula:





x= -b+/- Root of (b^2-4ac)


----------------------------------


2a





You just have to get the coefficients of the equation when it is in the general form:





ax^2 + bx+ c =0
Reply:Mo is right





you can see the steps at http://www.math-help.info/Proof_of_Quadr...
Reply:Solve this by completing the square.


ax^2 + bx = -c


x^2 + b/a * x = -c/a





x^2 + b/a * x + b^2/4a^2 = - c/a + b^2/4a^2





(x + b/2a)^2 = [b^2 - 4ac}/4a^2





and solving for x:





(x - b/2a) = +/- sqrt[b^2 - 4ac]/2a





x = -b/2a +/- sqrt[b^2 - 4ac]/2a
Reply:The first answer is correct, though a little brief.





axx + bx + c = 0


xx + bx/a + c/a = 0


xx + bx/a + bb/4aa - bb/4aa + c/a = 0


(x+b/2a)^2 - bb/4aa + c/a = 0


(x+b/2a)^2 = bb/4aa - c/a


(x+b/2a)^2 = (bb-4ac)/(4aa)


x + b/2a = (+/-) (bb-4ac)^0.5/(2a)


x = (-b +/- (bb-4ac)^0.5)/2a
Reply:You'd complete the square:





ax² + bx + c = 0





Subtract both sides by c and then divide both sides of that by a:





ax² + bx = -c


x² + (b/a)x = -c/a





To complete the square, you add to both halves of the equation the square of half the coefficient of x:





x² + (b/a)x + (b/2a)² = (b/2a)² - c/a





Factor the left half, perform the subtraction on the right half:





(x + (b/2a))² = b²/4a² - c/a


(x + (b/2a))² = b²/4a² - 4ac/4a²


(x + (b/2a))² = (b² - 4ac)/4a²





Now take the square root of both halves:





x + (b/2a) = ±√[(b² - 4ac)/4a²]





simplify





x = (-b/2a) ±√[(b² - 4ac)] / 2a


x = [-b ±√(b² - 4ac)] / 2a





That's how the quadratic equaton is derived.
Reply:check this out


http://www.csm.astate.edu/algebra/qform....